Final PDF file of OPT Class 9 2077

x 210° 225° 240° 270° 300° 315° 330° 360°

y = cos x –0.87 –0.71 –0.5 0 0.5 0.71 0.87 1

Y 90° 180° 270° 360°
90°
1.0 90° 180° 270° 360° X
0.8
0.6 180° 270° 360°
0.4
0.2

X'

-0.2
-0.4
-0.6
-0.8
-1.0

Y'

Example 6:

Draw the graph of y = tan x. Trigonometry

 As we have y = tan x

First, we list the values of the angle x ranging from 0º to 360º (i.e. 0 to 2pc) and the corresponding
values of y i.e. tan x.

x 0° 30° 45° 60° 70° 80° 90° 100° 110° 120° 135° 150° 180°

y = tan x 0 0.58 1 1.73 2.75 5.67 ∞ –5.67 –2.75 –1.73 –1 –0.58 0

x 210° 225° 240° 250° 260° 270° 280° 290° 300° 315° 330° 360°
y = tan x 0.58 1 1.73 2.75 5.67 ∞ –5.67 –2.75 –1.73 –1 –0.58 0

Y 90° 180° 270° 360°
90°
1.0 90° 180° 270° 360° X
0.8
0.6 180° 270° 360°
0.4
0.2

X'

-0.2
-0.4
-0.6
-0.8
-1.0

Y'

Trigonometric Ratios of Standard Angles ~ 197

EXERCISE : 5.6A

Find the value of: b. 2 3 sin 30º cos 45º tan 60º cosec 45º
1. a. 3 sin 45° cos 45º + 2 sec 60º cosec 90º d. sin330º – cos260º + tan345º
c. sin230º + cos245º + sin260º

2. a. tan2 p3c + 4 cos2 p4c + 3 sec2 p6c

b. cot2 p3c + 2 cos pc + sin2 p2c – cot pc + sin2 p3c
6 6

c. 3 tan2 pc + 4 cos2 p6c – sec2 p4c – 4 sin2 p3c
6

d. tasne2cp62cp4cc osseeccp3pc6csitnanp4cp6c

3. a. If q = 30°, then find sin2q.

b. If q = 60°, then find sin2q + cos2q.

c. If q = 30°, then prove that sin q = cos 2q.

4. Find the value of x:

a. x sin 30° = cos 60°

b. x sin 45°. cos 45° = 2 cos 90°

5. a. If q = 30°, then prove that sin 2q = 2 sin q. cos q

b. If q = 45° and a = 30°, prove that cos q + sin q sin a = 3
sin q 2

c. If q = 60° and b = 30°, then prove that tan(q + b) = tan q + tan b
1 – tan q tan b
6. Prove the following:

a. sin 30° + cos 60° = tan 30º b. 2 tan 30° = tan 60º
2 sin 60° 1 – tan230°

c. cos230º – sin230º = 1 – ttaann223300°° d. 11 +– ssiinn 3600°° = 2 + 3
2 +

7. a. tan2 pc + 2 sin2 pc + cos2 pc – tan2 pc = – 5
6 3 6 3 12

cos2 pc – sin2 pc = 1 – 2 sin2 p6c = cos pc
+ sin2 3
b. 6 6
cos2 pc pc

6 6

1 + cos pc pc pc 2
– cos 6 6
c. 6 = cosec + cot
1 pc

6

198 ~ Perfect Optional Mathematics Class 9

8. a. Find the height of the given tree. P

b. Calculate the value of q, from the given figure. 45° 15m R
A
P

c. Find the distance between A and C. q20m
40m
B 40m C
A Trigonometry

30°

BC

EXERCISE : 5.6B

Prove the following:

1. a. 3 sin260º + 4  cot230º – 2 cosec260º – 3  tan230º = 3 13
3 4

b. sin230º + cos260º + tan245º + sec260º – cosec230º = 1 21 

2. a. sin2 pc cosec pc – 2 sec2 pc + 3 cos pc tan pc + sin2 pc = 0
3 6 4 3 4 2

b. 3 tan2 pc + 4 cos2 p6c  – 1  sec2 p4c – 1  sin2 p6c = 3
6 3 2 3 4

c. tan2 p6c + 1  sin2 p4c + 1  cos2 p6c + 1 = 7
2 3 tan2 p3c 6

d. 34 tan2 p6c + 3 cosec2 p3c – 2 sin2 p3c – 3 cos2 p6c = 3
4 4

e. tan pc sin pc + sin pc cos pc + cos pc sin pc = 23
3 6 6 2 2 3

Trigonometric Ratios of Standard Angles ~ 199

i. tan2 pc + 4 sin pc + tan pc – 6 cot2p3c + cos2 p6c = 2 112
6 6 4

3. If A = 0º, B = 30º, C = 45º, D = 60º, E = 90º, prove that

a. tan (D – B) = tan D – tan B b. sin 3C = 3 sin C – 4 sin3C
1 + tan D tan B d. cos (B + D) = cos B cos D – sin B sin D

c. sin E + cot C + cos2B = 141

e. sin (E + B) sin (E – B) = sin2E – sin2B f. cos 2B = cos2B – sin2B

g. 11 +– tan B = 1 + sin D h. 13 + 7 +  8 sin B = 4
tan B 1 – sin B

4. Find the value of x from the following equations:

a. x sin 60º cos 60º tan 45º tan 30º = tan 60º sin 30º

b. x sin pc cos pc tan pc = tan2 p4c – cos2 p2c
4 4 6

c. sin2 23pc – cos2 23pc = x sin2 p4c cos2 p4c

d. 3 sin 60° + x cos 30° tan 45° = x cot 30°

e. 12x tan245° – 12 sin260° – 6 cot230° + 4 sec245° = 17

f. sin 30° + 2 cot230° + x cos230° = 8 + tan245° + cos 60°

g. x + 3 tan230° + 4 cos260° = sec245° + 4 sin260°

h. 2 cot pc = x sin p3c cos pc – tan pc
3 6

5. Draw the graph of the following trigonometric functions:

a. y = sin x (0º ≤ x ≤ 360º) b. y = cos x (0º ≤ x ≤ 360º)

c. y = tan x (0º ≤ x ≤ 360º) d. y = cosec x (0º ≤ x ≤ 360º)

e. y = 2 sin x (–180º ≤ x ≤ 180º) f. y = cos 3x (0º ≤ x ≤ 90º)

g. y = sin x + 3 (–90º ≤ x ≤ 90º) h. y = sin x + cos x (0º ≤ x ≤ 180º)



200 ~ Perfect Optional Mathematics Class 9

5.7 Compound Angles

Trigonometric Ratios of Compound Angles

The sum or the difference of two or more angles is said to be a Compound Angle. If A and B are two angles
then A + B and A – B are known as the compound angles. We express the trigonometrical ratios of the
compound angle A + B and A – B in terms of trigonometrical ratios of the angles A and B. The trigonometrical
ratios of the angles A + B and A – B are known as the addition and subtraction formula respectively.

Let XOX' and YOY' be rectangular X-axis and Y-axis respectively. Consider a unit circle (the circle having

radius 1 unit) with centre at origin O. Let the circle cuts the positive Y
P(x, y)
X-axis at M, the co-ordinates of M is (1, 0). If P(x, y) be any point on
the circumference, draw PN^OX such that POX = θ then we have

sin θ = PN = y1 i.e. y = sin θ X' q X
OP
O N M
cos θ = ON = 1x i.e. x = cos θ
OP

Hence, P(x, y) = P(cos θ, sin θ)

a. Cosine of Angle (A – B) Y'

Let P and Q are two points on the circumference of the circle such Y
that POX = A and QOX = B (A > B) as shown in figure. The P(cos A, sin A)
co-ordinates of points P and Q are (cos A, sin A) and (cos B, sin B)
respectively. Now, X' A Q(cos B, sin B) Trigonometry
B MX

O

PQ2 = [(cos B – cos A)2 + (sin B – sin A)]2

= cos2B – 2 cos A cos B + cos2A + sin2B Y'

– 2 sin A sin B + sin2A

= 2 – 2 (cos A cos B + sin A sin B) ... ... ... (i)

Rotate ®POQ through angle B about origin in the clockwise direction Y
so that OQ coincides with OM and ®POM = A – B. The co-ordinates
of point P is [cos (A – B), sin (A – B)] and that of Q is (1, 0). Now, P

PQ2 = [1 – cos (A – B)]2 + [0 – sin (A – B)]2 X' O A – B X

QM

= 1 – 2 cos (A – B) + cos2(A – B) + sin2(A – B)

= 2 – 2 cos (A – B) ... ... ... (ii) Y'

From (i) and (ii), we have,

cos (A – B) = cos A cos B + sin A sin B

Compound Angles ~ 201

b. Cosine of Angle (A + B)

Let P and Q are two points on the circumference of the circle, Y
such that ®POX = A and ®QOX = – B. The co-ordinates of P and
P
Q are (cos A, sin A) and [cos (– B), sin (– B)] or (cos B, – sin B)
X' A MX
respectively, then Q
O – B
PQ2 = (cos B – cos A)2 + (– sin B – sin A)2

= cos2B – 2 cos A cos B + cos2A + sin2B

+ 2 sin A sin B + sin2A Y'

= 2 – 2 (cos A cos B – sin A sin B) ... ... ... (i)

Now, rotate OQ through angle B about O in anticlockwise direction Y
so that OQ coincides with OM and ®POM = A + B. The co-ordinates P

of point P is [cos (A + B), sin (A + B)] and that of Q is (1, 0).

Now, PQ2 = [1 – cos (A + B)]2 + [0 – sin (A + B)]2 X' O A + B X
= 1 – 2 cos (A + B) + cos2(A + B) + sin2(A + B)
QM

= 2 – 2 cos (A + B) .. ... ... (ii)

From (i) and (ii), we have Y'

cos (A + B) = cos A cos B – sin A sin B

Note: This relation can be obtained by replacing B by – B in above relation.

i.e. cos [A – (– B)] = cos A cos (– B) + sin A sin (– B) = cos A cos B – sin A sin B.

c. Sine of Angle (A – B)

Let P and Q are two points on the circumference of the circle such Y

that ®POX = pc – A and ®QOX = – B. P
2
The co-ordinates of points P and Q are (sin A, cos A) and [cos (–  B),
p2 – A
sin (–  B)] or (cos B, – sin B) respectively. Now, X' O – B M X

PQ2 = (cos B – sin A)2 + (– sin B – cos A)2 Q

= cos2B – 2 cos B sin A + sin2A + sin2B

+ 2 sin B cos A + cos2A Y'

= 2 – 2 (sin A cos B – cos A sin B) ... ... ... (i)

Rotate ®POQ through angle B about origin in the anticlockwise Y
P
direction so that ®POM = pc – A + B.
2

The co-ordinates of point P is p2 –(A–B)

cos pc – (A – B) , sin pc – (A – B) X' O QM X
2 2

i.e. [sin (A – B), cos (A – B)] and that of Q is (1, 0). Now,

PQ2 = [1 – sin (A – B)]2 + [0 – cos(A – B)]2 Y'

= 1 – 2 sin (A – B) + sin2(A – B) + cos2(A – B)

= 2 – 2 sin (A – B) ... ... ... (ii)

202 ~ Perfect Optional Mathematics Class 9

From (i) and (ii), we have

sin (A – B) = sin A cos B – cos A sin B

Note: This relation can be obtained by replacing A by pc + A in above relation
2
cos (A – B) = cos A cos B + sin A sin B.
pc pc pc
i.e. cos 2 + A – B = cos 2 + A cos B + sin 2 + A sin B

i.e. – sin (A – B) = – sin A cos B + cos A sin B

i.e. sin (A – B) = sin A cos B – cos A sin B

Sine of Angle (A + B)

Let P and Q are two points on the circumference of the circle such Y
P
that ®POX = pc – A and ®QOX = B.
2 Q

The co-ordinates of points P and Q are X' B p2 – A X

cos pc + A , sin pc + A OM
2 2

i.e. (sin A, cos A) and (cos B, sin B) respectively. Now, Y'

PQ2 = (cos B – sin A)2 + (sin B – cos A)2

= cos2B – 2 cos B sin A + sin2A + sin2B – 2 sin B cos A + cos2A

= 2 – 2 (sin A cos B + cos A sin B) ... ... ... (i)

Rotate ®POQ through angle B about origin in clockwise direction so Y Trigonometry

that OQ coincides with OM and ®POM = pc – A – B. P
2

The co-ordinates of P is cos pc – (A + B) , sin pc – (A + B) X' O p2 – (A + B) X
2 2
QM

i.e. [sin (A + B), cos (A + B)] and that of Q is (1, 0).

Now, PQ2 = [1 – sin (A + B)]2 + [0 – cos (A + B)]2

= 1 – 2 sin (A + B) + sin2(A + B) + cos2(A + B) Y'

= 2 – 2 sin (A + B) ... ... ... (ii)

From (i) and (ii), we have

sin (A + B) = sin A cos B + cos A sin B

Note: This relation can be obtained by replacing A by pc – A in above relation
2
i.e. cos (A – B) = cos A cos B + sin Asin B.

i.e. cos pc – A – B = cos pc – A cos B + sin pc – A sin B.
2 2 2

i.e. sin (A + B) = sin A cos B + cos A sin B.

Now, using the formula of sin (A + B), cos (A + B), sin (A – B), cos (A – B), we find the formula of
tan (A + B), tan (A – B), cot (A + B) and cot (A – B).

tan (A + B) = sin (A + B) = sin A cos B + cos A sin B
cos (A + B) cos A cos B – sin A sin B

Compound Angles ~ 203

sin A cos B + cos A sin B tan A + tan B
cos A cos B cos A cos B 1 – tan A tan B
= =
cos A cos B sin A sin B
cos A cos B – cos A cos B

∴ tan (A + B) = tan A + tan B
1– tan A tan B

Similarly, tan (A – B) = tan A– tan B B
1+ tan A tan

cot (A + B) = cot A cot B – 1 and
cot B + cot A

cot (A – B) = cot A cot B + 1
cot B – cot A

Again, sin (A + B) sin (A – B)

= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)

= sin2A cos2B – cos2A sin2B

= sin2A (1 – sin2B) – (1 – sin2A) sin2B

= sin2A – sin2B

= 1 – cos2A – 1 + cos2B

= cos2B – cos2A.

∴ sin (A + B) sin (A – B) = sin2A – sin2B = cos2B – cos2A

Similarly, cos (A + B) cos (A – B) = cos2A – sin2B = cos2B – sin2A
Example 1:

Find the value of: a. cos 75º
cos 15º + sin 15º
b. tan 15º c.

 a. As, 75º = 45º + 30º

cos 75º = cos (45º + 30º)

= cos 45º cos 30º – sin 45º sin 30º

= 1 × 3 – 1 × 1 = 3–1
2 2 2 2 2  2

b. As, 15º = 45º – 30º

tan 15º = tan (45º – 30º)

= tan 45º + tan 30º
1 – tan A tan B

1– 1 3–1
3 3
= ×1 =
3+1
1 + 1 3 3

204 ~ Perfect Optional Mathematics Class 9

= 3 – 1 × 3–1
3 + 1 3–1

= 3  2 – 2 3 + 1
3–1

= 4 –2 3=2– 3
2

c. cos 15º + sin 15º

= cos (45º – 30º) + sin (45º – 30º)

= cos 45º cos 30º + sin 45º sin 30º + sin 45º cos 30º – cos 45º sin 30º

= 1 × 3 + 1 × 1 + 1 × 3 – 1 × 1
2 2 2 2 2 2 2 2

= 3 + 3 = 2  3 = 3
2  2 2  2 2  2 2

Example 2:

Prove that:

a. cos 36º 25' cos 23º 35' – sin 36º 25' sin 23º 35' = 1
2

b. cos 14° – sin 14° = cot 59º c. cos 18º – sin 18º = 2 sin 27º
cos 14 + sin 14°

 a. LHS = cos 36º 25' cos 23º 35' – sin 36º 25' sin 23º 35' Trigonometry

= cos (36º25' + 23º35')

= cos 60º = 1 (RHS)
2

b. RHS = cot 59°

= cot (45° + 14°) = cot 45° cot 14° – 1
cot 45° + cot 14°

cot 14° –1 cos 14° – 1
1+ cot 14° sin 14°
= =
cos 14°
sin 14° + 1

cos 14° – sin 14°

= cos sin 14° 14° = cos 14° – sin 14° (LHS)
14° + sin cos 14° + sin 14°

sin 14°

c. LHS = cos 18º – sin 18º

= cos (45º – 27º) – sin (45º – 27º)

= cos 45º cos 27º + sin 45º sin 27º – sin 45º cos 27º + cos 45º sin 27º

= 1 (cos 27º + sin 27º – cos 27º + sin 27º)
2

= 1 × 2 sin 27º = 2 sin 27º (RHS).
2

Compound Angles ~ 205

Example 3:

If sin α = 1 and cos β = 25, find the value of cos (α + β) and sin (α + β).
10

 Here, sin α = 1 and cos β = 2
10 5

cos α = 1 – sin2a = 1 – 1 = 9 = 3
10 10 10

sin β = 1 – cos2b = 1 – 4 = 1 = 1
5 5 5

Now, cos (α + β) = cos α cos β – sin α sin β

= 130 × 2 – 1 × 1
5 10 5

= 160–×1 5 = 5 = 1
52 2

and sin (α + β) = sin α cos β + cos α sin β

= 1 × 2 + 3 × 1
10 5 10 5

= 2 + 3 = 5 = 1
10 × 5 5 5 2

Example 4:

Prove that:

a. 1 + tan A tan 2A = sec 2A

b. sin A + sin (A + 120°) + sin (A – 120º) = 0

 a. LHS = 1 + tan A tan 2A

= 1 + sin A sin 2A
cos A cos 2A

= cos A cos 2A + sin A sin 2A
cos A cos 2A

= cos (2A – A) = cos cos A 2A = sec 2A. (RHS)
cos A cos 2A A cos

b. LHS = sin A + sin (A + 120°) + sin (A – 120º)

= sin A + sin A cos 120º + cos A sin 120º + sin A cos 120º – cos A sin120°

= sin A + 2 sin A cos 120º

= sin A + 2 sin A × – 1
2

= sin A – sin A

= 0 (RHS)

206 ~ Perfect Optional Mathematics Class 9

Example 5:

Prove that:
a. tan 25º + tan 20º + tan 20º tan 25º = 1
b. tan 10º + tan 70º + tan 100º = tan 10º tan 70º tan 100º
c. tan 50º = tan 40º + 2 tan 10º

 a. As, 25º + 20º = 45º

tan (25º + 20) = tan 45º
or, 1ta–nta2n5°2+5°tatann2200°° = 1
or, tan 25º + tan 20º = 1 – tan 25º tan 20º

∴ tan 25º + tan 20° + tan 25° tan 20° = 1

b. LHS = tan 10º + tan 70º + tan 100º

= tan 10° + tan 70° + tan (180° – 80°)

= tan 10° + tan 70° – tan 80°

= tan 10° + tan 70° – tan (10° + 70°)

= (tan 10° + tan 70°) – (tan 10° + tan 70°)
1 – tan 10° tan 70°
Trigonometry
= (tan 10° + tan 70°) 1 – 1 – tan 1 tan 70°
10°

= (tan 10° + tan 70°) 1 – tan 10° tan 70° – 1
1 – tan 10° tan 70°

= tan 10° + tan 70° × – tan 10° tan 70°
1 – tan 10° tan 70°

= tan (10° + 70°) × – tan 10° tan 70°

= – tan 80° tan 10° tan 70°

= tan (180° – 80°) tan 10° tan 70°

= tan 10° tan 70° tan 100° (RHS)
Alternately,

We have, 10° + 70° = 80°

tan (10° + 70°) = tan 80°

or, 1ta–nta1n0°1+0°tatann7700°° = tan (180° – 100°)

or, tan 10° + tan 70° = – tan 100°
1 – tan 10° tan 70°

or, tan 10° + tan 70° = – tan 100° + tan 10° tan 70° tan 100°

Compound Angles ~ 207

∴ tan 10° + tan 70° + tan 100° = tan 10° tan 70° tan 100°
c. As, 50° – 40° = 10°
tan (50° – 40°) = tan 10°

or, 1ta+nta5n0°5–0°tatann4400°° = tan 10°
tan 50° – tan 40°
or, 1 + tan 50° tan (90° – 50°) = tan 10°

or, 1ta+nta5n0°5–0°tacnot4500°° = tan 10°

or, tan 50° – tan 40° = tan 10°
1 + 1

or, tan 50° – tan 40° = 2 tan 10°
∴ tan 50° = tan 40° + 2 tan 10°

Example 6:

If tan α = 5 and tan β = 111, prove that α + β = 45º.
6

 We have, tan α = 5 and tan β = 111.
6
tan a + tan b
tan (α + β) = 1 – tan a tan b

5 + 1 55 + 6
6 11 66
= =
– 5 1 66 – 5
1 6 × 11 66

= 61 × 66 = 1 = tan 45°
66 61

∴ α + β = 45°.

Example 7:

An angle θ is divided into two parts so that the ratio of the tangents of the parts is k. If α be the
k – 1
difference between the two parts then show that sin α = k + 1 sin θ.

 Let θ be divided into two parts A and B so that A + B = θ, A – B = α and from equation tan A : tan B = k.

or, ttaann A = k
B 1

By componendo and dividendo,

or, ttaann A – tan B = k – 1
A + tan B k + 1

sin A + sin B k – 1
cos A cos B k + 1
or, =
sin A sin B
cos A + cos B

208 ~ Perfect Optional Mathematics Class 9

or, ssiinn A cos B – cos A sin B = k – 1
A cos B + cos A cos B k + 1

or, ssiinn (A – B) = k – 1
(A + B) k + 1

or, ssiinn a = k – 1
q k + 1

∴ sin α = k – 1 sin θ.
k + 1
EXERCISE : 5.7A

1. Find the values of: b. sin 15° c. tan 15° d. cot 15°
a. cos 15° f. tan 105° g. sin 75° h. cos 75°
e. sin 105°
2. Prove that:

a. sin 19°40' cos 25°20' + cos 19°40' sin 25°20' = 1
2
1
b. cos 13°25' cos 58°25' + sin 58°25' sin 13°25' = 2

c. cos θ cos (φ – θ) – sin θ sin (φ – θ) = cos φ

d. sin (x + 1)A sin (x + 2)A + cos (x + 1)A cos (x + 2)A = cos A

e. cos (45° – A) (cos 45° – B) – sin (45° – A) sin (45° – B) = sin (A + B) Trigonometry

f. sin (45° + A) cos (45° – B) + cos (45° + A) sin (45° – B) = cos (A – B)

g. sin (60° – θ) cos (30° + θ) + cos (60° – θ) sin (30° + θ) = 1

3. Prove that:

a. sin (A + B) + sin (A – B) = 2 sin A cos B b. sin (A + B) – sin (A – B) = 2 cos A sin B

c. cos (A + B) + cos (A – B) = 2 cos A cos B d. cos (A – B) – cos (A + B) = 2 sin A sin B

e. sin (A + B) sin (A – B) = sin2A – sin2B f. cos (A + B) cos (A – B) = cos2A – sin2B

4. Prove the following identities:

a. sceocs2qq – sin q = cos 3θ b. ttaannAA –+ ttaann BB = sin (A + B)
cosec 2q sin (A – B)

c. tan2A – tan2B = sin (A + B) sin (A – B) d tan (A + B) + tan (A – B) = sin 2A
cos2A cos2B cos2A – sin2B

e. tan (A + B) tan (A – B) = sin2A – ssiinn22BB f. 1 + tan 2A tan A = 1
cos2A – sec 2A

g. 1 – tan A tan B = sec A sec B
sec (A + B)

5. Prove the following identities:

a. cos (θ – 45°) = 1 (cos θ + sin θ) b. sin (θ – 45°) = 1 (sin θ – cos θ)
2 2

c. sin (45° + θ) – cos (45° – θ) = 0 d. sin (45º + A) + cos (45º + A) = 2 cos A

Compound Angles ~ 209

e. sin (45º + A) sin (45° – A) = 1 (cos2A – sin2A)
2

f. 2 sin (45° + θ) sin (45° – θ) = cos2θ – sin2θ

g. 2 cos (45° + θ) cos (45° – θ) = cos2θ – sin2θ

6. Prove the following identities:

a. tan 75º + cot 75° = 4 b. sin 105° + cos 75° = 3
2

c. sin 75° – cos 75° = 1 d. cos 105° + cos 15° = 1
2 2

7. Without using calculator or table, prove the following :

a. cos 15° – cos 75° = 1 b. sin 75° + cos 75° = 3
2 2

c. sin 105° + cos 15° = 3 + 1 d. sin 15° + cos 15° = 3
2 2

e. cos 15° – sin 15° = 1 f. tan 15° + cot 15° = 4
2

8. Show that:

a. cos 55° + cos 35° = 2 cos 10° b. sin 37° + cos 37° = 2 cos 8°

c. cos 18° – sin 18° = 2 sin 27° d. cos 65° + sin 65° = 2 cos 20°
e. sin 65° – sin 25° = 2 sin 20° f. cos 20° – sin 20° = 2 sin 25°
9. Prove the following identities:

a. 1ta–nta2n2°2+2°tatann3388°° = 3 b. 1ta+nta5na5–atatann44aa = tan α

c. 1ta+nta(an + b) – tan b = tan α d. tan 10° + tan 35° + tan 10° tan 35° = 1
(a + b) tan b

e. tan 10° + tan 40° + tan 130° = tan 10° tan 40° tan 130°

f. tan 20° + tan 72° + tan 88° = tan 20° tan 72° tan 88°

g. tan 5A – tan 3A – tan 2A = tan 5A tan 3A tan 2A

h. tan (α – β) tan (α + β) = sin2a – sin2b
cos2b – sin2a

10. Prove that :

a. (1 + tan 24°) (1 + tan 21°) (1 + tan 23°) (1 + tan 22°) = 4

b. (1 + cot 55°) (1 + cot 80°) (1 + cot 77°) (1 + cot 58°) = 4

11. Prove that :

a. cos A + cos (120° + A) + cos (120° – A) = 0

b. sin α + sin a + 2pc + sin a + 4pc =0
3 3

210 ~ Perfect Optional Mathematics Class 9

EXERCISE : 5.7B

1. Prove that:

a. ccooss 35° – sin 35° = tan 10° b. sin 70° – cos 70° = tan 25°
35° + sin 35° sin 70° + cos 70°

c. ccooss 11° – sin 11° = cot 56° d. sin 38° – cos 38° = – tan 7°
11° + sin 11° cos 38° + sin 38°

e. ccooss 10° – sin 10° = sin 35° sec 35° f. cos 17° – sin 17° = cos 62° cosec 62°
10° + sin 10° cos 17° + sin 17°

2. Prove that:

a. tan 70° = tan 20° + 2 tan 50° b. tan 80° = tan 10° + 2 tan 70°

3. Prove that:

a. cos (A + B) cos C – cos (B + C) cos A = sin B sin (C – A)

b. cos (B + C) cos (B – C) – cos (C + A) cos (C – A) = sin (A + B) sin (A – B)
pc pc
tan 4 + q + tan 4 – q
cosec 2θ = – tan
c. pc pc
tan 4 4
+ q – q

d. csoins (AAc–osBB) + sin (B – C) + sin (C – A) = 0
cos B cos C cos C cos A

4. a. If sin α = 3 and cos β = 1123, find the values of sin (α + β) and tan (α + β).
5

b. If sin A = cos B = 1 , find the values of sin (A + B) and tan (A – B). Trigonometry
2

5. a. If tan A = m and tan B = m1 , prove that A + B = p2c.

b. If sin θ = 1 and sin f = 1, show that θ + f = p4c.
10 5

c. If cos A = 4 and cos B = 7 , prove that A + B = p4c.
5 52

6. If A + B = p4c, prove the following:

a. (1 + tan A) (1 + tan B) = 2 b. (cot A – 1) (cot B – 1) = 2

c. tan A + tan B + tan A tan B = 1

7. a. If sin (A + B) = 2 sin (A – B), prove that tan A = 3 tan B.

b. If tan α = k tan β, prove that (k + 1) sin (α – β) = (k – 1) sin (α + β).

8. Express the following in terms of the trigonometrical ratios of A, B and C.

a. sin (A + B + C) b. tan (A + B + C)

9. Prove that:

a. tan 1 tan A – cot 1 cot A = cot 4A b. tan 1 tan A – cot 1 cot A = cot A
3A + 3A + 2A – 2A –

c. tan 1 tan A – cot 1 cot A = cot 2A d. cot cot A 3A – tan tan A A = 1
3A – 3A – A – cot 3A – tan



Compound Angles ~ 211

6 VECTOR

Curriculum Contents Included Periods

- Introduction of vector and types of vectors - column, row, position, unit, zero, 17
equal and negative vectors

- Like and unlike vecotors
- Magnitude and direction of vector
- Operation of vectors
- Multiplication of a vector by a scalar (condition of parallelism)
- Addition and subtraction of vectors
- Laws of vector addtion

Objectives

On completion of this unit, students will be able to:
define vectors and state the representation of vectors.

plot the given vectors on graph and find the components of the vectors represented on the graph.

find the direction and magnitude of the vector and state the types of vectors.

perform the operations of vectors (Addition and Subtraction)

multiply the vector by the scalar and state the condition of parallelism of vectors.

Materials Parallelogram Law of Vector Addition
Triangle Law of Vector Addition
CD
C

AB AB
→AB + →BC = →AC A→D = →AB + →AC

The triangular law of vector addition states that The parallelogram law of vector addition states
"If the magnitude and direction of two vectors that "If two adjacent sides of a parallelogram
are represented by two sides of a triangle taken in through a point represents two vectors in
order. The magnitude and direction of their sum magnitude and direction, their sum is given by the
is given by the third side taken in reverse order." diagonal of the parallelogram through the same

point in magnitude and direction."

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Vectors 1 mark each 5 marks each Questions Marks
2 marks each 4 marks each
1 1 4 10
2–

212 ~ Perfect Optional Mathematics Class 9

6.1 Vectors and its Components

Introduction

The quantities which can be measured are called physical quantities. Some physical quantities have
magnitudes only but some physical quantities have magnitudes as well as directions.

Let us study the addition of some physical quantities. For example, if we have two rectangles of areas 6cm2
and 8cm2, the total areas of these two rectangles is (6 + 8)cm2 i.e. 14cm2.

+ =

6 cm2 8 cm2 14 cm2

If two forces 4N and 5N act on a body. What is the total force acting on the body. 9N or 1N ? Its answer is

uncertain unless their direction are known. If two forces have same direction, the total force on the body is

4N + 5N = 9N. 4N 9N
5N

If two forces have opposite direction, the total force on the body is 5N – 4N = 1N.

5N 4N
1N + 4N 4N

In the calculation of some physical quantities, the direction has very important role. Vector

Scalars and Vectors

A physical quantity which can be measured with its magnitude only is known as scalar quantity. For

example, mass, length, volume, area etc. A physical quantity which can be measured with its magnitude as

well as direction is known as vector. For example, force, velocity, acceleration, weight etc.

Representation of a Vector

A vector is generally represented by a directed line segment. The length B

of the line segment is chosen according to some scale to represent the

magnitude and the arrow to represent the direction of the vector. A A
vector from point A to point B is denoted by →AB .

Vectors and its Components ~ 213

We call the point A, the initial (starting) point or tail and the point B, Q
the terminal (end) point or head of the vector →AB . A vector is also

represented by components form.Avector from point Pto Q is denoted by PM
x PM
y = MQ , where PM = x is called x-component and MQ = y is

called y-component.

Directed Line Segments

"If two points of a line are taken, one as the initial point and another as the terminal point, then it is called

a directed line segment."

If M→N is a directed line segment, then  M→N  represents the length and N

arrow (→) shows the direction (i.e. from M to N).

The two ends of di rected line segments are not interchangeable. M B
So the directed line segments →AB and →BA are not the same. These A
A
directed line segments are equal in magnitude but opposite in direction. B
So, →AB = – →BA .

If the length and direction of two directed line segments are same they are called equivalent directed line

segments or equal directed line segments.

a. →CD and →PQ have the same direction. D

b. →CD and →PQ have the same length. C Q
Q
Hence, →CD = →PQ . P
P
In the figure, the terminal point of →PQ and initial point of →QR are the
same. The initial point of →PQ + →QR is P and its terminal point is R.

Hence, →PQ + →QR = →PR R

If →AB and →BC are two equal directed line segments, AB C
i.e. →AB = →BC

Here, →AC = →AB + →BC

or, →AC = 2 →AB

Here, directed line segment →AC is expressed in terms of →AB .

In general, if k be any real number and →AB be a directed line segment, then k →AB represent another directed
line segment →AC such that.

214 ~ Perfect Optional Mathematics Class 9

a. C lies in the direction of →AB if k > 0 and  →AC   = k  →AB  
b. C lies in the opposite direction of →AB if k < 0 and  →AC   = – k  →AB  

Position Vector

A point P(x, y) determines its position with reference to origin O(0, 0). Join OP, draw PN⊥OX, so ON = x

and PN = y. Y
P(x, y)
To displace from O to N and again from N to P is same as to displace

from O to P. This horizontal displacement O→N with ON = x and vertical y

displacement →NP with NP = y combinely gives the displacement →OP . X' O xN X

The displacement →OP is written as (x, y) or x . Y'
y
Here, →OP is said to be the position vector of P. Here x and y are said

to be x– component (horizontal component) and y– component (vertical component) of →OP respectively.

The position vector of point P(x, y) is O→P = x
y
Vectors in Terms of Components

Let P(x1, y1) and Q(x2, y2) be two points on the plane. From P and Q Y

draw PM⊥OX and QN⊥OX and again draw PR⊥QN. Q
R
Here, PR = MN = ON – OM = x2 – x1 and The displacement →PQ has P
QR = QN – RN = QN – PM = y2 – y1. NX
horizontal displacement and vertical displacement →RQ . OM
X' Y'

Hence, →PQ = (x2 – x1, y2 – y1) or x2 – x1
y2 – y1
Here, we have position vectors of P and Q are →OP = x1 and O→Q = x2
y1 y2 Vector
→PQ = x2 – x1 x2 x1 = O→Q – →OP .
So, y2 – y1 = y2 – y1

∴ →PQ = O→Q – →OP x2 – x1
The vector joining two points P(x1, y1) & Q(x2, y2) is →PQ = O→Q – →OP = y2 – y1

Example 1:

Find the vectors represented by the following directed line segments in component form.

 For vector →AB

starting from A, horizontal component (right) = 3 and vertical B
D
component (up) = 1. Hence, →AB = 3
1 A
For vector →CD , starting from C horizontal component (left) = – 3 and PM C

vertical component (up) = 2. Hence, →CD = – 3 Q
2 N

Vectors and its Components ~ 215

For vector →PQ ,
starting from P, horizontal component (left) = – 2 and vertical component (below) = – 2.

Hence, →PQ = – 2
– 2

For vector M→N,

starting from M, horizontal component (right) = 2 and vertical component (below) = – 3

Hence, M→N = 2
– 3

Example 2:

Represent the following vectors by the directed line segments.

a. →AB = 2 b. →CD = – 2 c. →PQ = – 4 d. M→N = 4
– 2 – 3 2 2
 a.
→AB = 2 A
– 2

Horizontal component = + 2, Vertical component = – 2 B

Starting from A, 2 units right and then 2 units down will give the point B. Then the directed line

segment →AB will give 2 .
– 2

∴ Similarly, the directed line segments of vectors

b. →CD = – 2 c. →PQ = – 4 and d. M→N = 4
– 3 2 2

and can be drawn as shown in figure.

C
N

Q

DM
P

EXERCISE : 6.1

1. a. Define position vector of a point with an example.

b. Give two examples of vectors and scalar each.

2. a. What do m and n represent in →a = m  ?
n

b. Two points A(x1, y1) and B(x2, y2). Write the formula to calculate column vector →AB .

216 ~ Perfect Optional Mathematics Class 9

3. Find the vectors represented by the following directed line segments.

→d →e →g
→a →c →f

→b

4. From the figure given below, find the components of the following B
C
vectors. A

a. →AB b. →AC c. D→G

d. →EG e. →FA f. →GC GD

E
F

5. Find the vectors represented by the directed line segments joining the following points:

a. A(2, 5) and B(1, 6) b. P(– 4, 8) and Q(3, – 4)

c. M(4, – 6) and N(0, 3) d. C(3, 4) and D(– 5, – 3)

6. Represent the following vectors by the directed line segments.

a. →AB = 4 b. →PQ = 6
– 3 – 3

c. M→N =  –  7 d. →CD = 8
– 3 2

7. a. If A(4, 4), B(x, y) and →AB = 8 , find the values of x and y.
4

b. If A(a, 4), B(12, b) and →AB = 8 , find the values of a and b.
4

c. If A(6, 6), B(x, y) and →AB = 12 , find the values of x and y. Vector
6

d. If P(a, 6), Q(18, b) and →PQ = 12 , find the values of a and b.
6

8. a. If →AB = 6 and the co-ordinates of A is (– 2, – 2), find the c o-ordinates of B.
3

b. If →AB = 12 and the co-ordinates of A is (– 4, – 4), find the co-ordinates of B.
6

c. If →PQ = 12 and the co-ordinates of P is (– 6, – 6), find the co-ordinates of Q.
9

d. If →PQ = 24 and the co-ordinates of P is (– 8, – 8), find the co-ordinates of Q.
12



Vectors and its Components ~ 217

6.2 Magnitude and Direction of a Vector

Magnitude of a Vector

Magnitude of a vector is the length of its directed line segment from the initial point to the terminal point.

The modulus or the magnitude of vector →OP is denoted by  →OP   .

If →OP = position vector of P = x
y
Draw PM⊥OX, then
x– component of →OP = OM = x Y
P(x, y)

y– component of →OP = MP = y y

Then, OP2 = OM2 + MP2 X' O q M X
or, OP2 = x2 + y2 x

or, OP = x2 – y2 Y'

Hence, OP =  →OP   = x2 – y2

In general, if a vector →AB displaces a point A(x1, y1) to the point B(x2, y2) then vector →AB is represented by

→AB = x2 – x1 Y B (x2, y2)
y2 – y1
(y2 – y1)
x– component (x) = x2 – x1

y– component (y) = y2 – y1 A (x2 – x1) R
Magnitude of →AB is
(x1, y1)
 →AB   = (x2 – x1)2 + (y2 – y1)2
X' O M N X

Y'

= (x-component)2 + (y-component)2

Hence, the magnitude of a vector is given by magnitude = (x-component)2 + (y-component)2

Direction of a Vector

Direction of a vector is measured as the angle made by the vector with positive direction of X-axis in
anticlockwise direction.

Let the position vector of point P is x and →OP makes angle q with Y
y P(x, y)

positive direction of X-axis in anticlockwise direction as shown. Draw y
PM⊥OX, then OM = x, PM = y and from ∆OPM,

tan θ = y X' O q M X
x x

∴ θ = tan–1 x Y'
y

218 ~ Perfect Optional Mathematics Class 9

For the direction of a vector, we find the tangent of the angle (θ) made by the given vector with the positive

X– axis.

For vector →OA = 3 , if q be the angle made by →OA with positive Y
2

X– axis A(3, 2)

tan θ = AN = 2 = 0.666 X' O (0, 0) N X
ON 3

or, tan θ = tan 34° (app.)

∴ direction of →OA (θ) = 34° Y'
Y
If A(x1, y1) and B(x2, y2) are the initial and terminal points of a vector B (x2, y2)

→AB then, →AB = x2 – x1 . Suppose →AB makes an angle q with positive (y2 – y1)
y2 – y1
q
direction of X-axis in anticlockwise direction. A (x2 – x1) R

(x1, y1)

From figure, BAR = θ and X' O M N X

tan θ = y2 – y1 Y'
x2 – x1

i.e. θ = tan– 1 y2 – y1 .
x2 – x1

If A(x1, y1) and B(x2, y2) are two points and vector →AB = x2 – x1 makes an angle θ with positive direction
y2 – y1

of X-axis in anticlockwise direction, the direction θ of vector →AB is given by

tan θ = y2 – y1 = y-component
x2 – x1 x-component
Vector
Note: a. If x-component = +ve and Y
y-component = +ve A
then the direction of vector is θ.
(+)

X' O q X

(+) N

Y'

b. If x-component = – ve and Y
y-component = +ve then the A
direction of vector is (180° – θ). (why?).
(+) 180° – q

X' q X

N (–) O

Y'

Magnitude and Direction of a Vector ~ 219

c. If x-component = – ve and X' N (–) Y
y-component = – ve then the direction q
of vector is (180° + θ) (why?) (–) 180° + q

OX

A
Y'

d. If x– component = +ve and Y
y-component = – ve then the direction
of vector is (360° – θ) (why ?) 360° – q (+) N X

X' O q

(–)

A
Y'

e. If x-component = +ve and y-component = 0 then the direction of vector is 0°.
f. If x-component = 0 and y-component = +ve then the direction of vector is 90°.
g. If x-component = – ve and y-component = 0 then the direction of vector is 180°.
h. If x-component = 0 and y-component = – ve then the direction of vector is 270°.

Example 1:

If →CD displaces the point C(10, 13) into D(7, 8),
a. Find the components of →CD
b. Find the magnitude of →CD and
c. Find the direction of →CD .

 a. x-component of →CD

= x2 – x1
= 7 – 10

= – 3

y-component of →CD

= x2 – x1
= 7 – 10

= – 3

Hence, →CD = – 3
– 5

220 ~ Perfect Optional Mathematics Class 9

b. Magnitude of →CD
=  →CD  

= (– 3)2 + (– 5)2

= 9 + 25

= 34 units
c. If q be the angle made by →CD with positive X– axis,

then tan θ = y-component = – 5
x-component – 3

or, tan θ = 1.666

or, tan θ = tan 59°

∴ θ = 59° (appr.)

Since, both the components of →CD are negative, the direction of →CD is 180° + 59°

= 239°

EXERCISE : 6.2

1. a. What is the length of vector →a = 4  ?
0

b. If →a = (a, b), find the magnitude of →a .

c. If A(6, 1) and B(5, 2), find the magnitude of →AB . Vector

d. If →AB = 23 , find the direction of →AB .
2

e. If A(5, 1) and B(6, 2), find the direction of →AB .

2. If →PQ displaces the point P(2, 8) to Q(– 2, 3)

a. Find the vector →PQ in the form of column vector.

b. Find the magnitude of →PQ .

c. Find the direction of →PQ .

Magnitude and Direction of a Vector ~ 221

3. Find the magnitude and direction of the following vectors.

3

a. 4 b. 12 c. 5 d. – 6
3 – 5 4 3

5

4. If →OA = 3 , →OB = –2 1 , →OC = 1 , O→D = – 7 and →OE = – 3 , find the folloiwng vectors.
5 – 3 0 – 7

a. →AB b. →BC c. →CD d. →ED

5. If →OA = 2 and →OB = – 1 be the position vectors of two points A and B respectively, find
– 3 5

a. →AB b.  →AB   c. direction of →AB

6. a. If →PQ displaces the point P(2, 4) to Q(– 2, 8) and →AB displaces the point A(3, – 4) to B(– 1, 0),
then prove that →PQ = →AB .

b. If →AB displaces the point A(2, 5) to B(7, 3) and →PQ displaces the point P(7, – 3) to Q(2, – 1),
then prove that →AB = – →PQ .

c. If M→N displaces the point M(– 4, – 2) to N(3, – 6) and →CD displaces the point C(7, 3) to D(a, b).
If M→N = →CD , find the values of a and b.

7. Find the values of x and y in each of the following conditions:
a. P(x, y), Q(16, 12), R(4, 4), S(12, 8) and →PQ = →RS .
b. A(x, 12), B(8, 2), C(6, y), D(12, 2) and →AB = →CD .
c. A(2, 4), B(6, 12), C(8, 0), D(x, y) and →AB = →CD .

8. a. A vector is a . Its angle with x-axis 45° and the magnitude is 2 2 . Find the vector.
b

b. A vector is x . Its magnitude is 5 and angle with x-axis is tan-1 3 . Find the values of x and y.
y 4



222 ~ Perfect Optional Mathematics Class 9

6.3 Types of Vectors

Row Vector and Column Vector

Let →AB = →a be a vector which displaces a point A(x1, y1) into B(x2, y2), then the x-component of →a is
x2 – x1 and its y-component is y2 – y1.
If x2 – x1 = x and y2 – y1 = y (say). Here, the vector →AB can be expressed
in two forms row and column. Y P(x2, y2)

→AB = →a = (x2 – x1 , y2 – y1) = (x, y) A y2 – y1
∴ →a = (x, y) is called a row vector and
(x1, y1) x2 – x1 M

→AB x2 – x1 x X' O X
y2 – y1 y
= →a = = Y'

∴ →a = x is called a column vector.
y

Hence, the row vector and the column vector are the method of representing a vector in a row or in a column form.

Equal Vectors

Two vectors are said to be equal if they have same direction and same C
magnitude. In other words, two vectors are said to be equal if their A
corresponding components are equal.
D
The vectors →AB and →CD in the figure are equal because B

→AB = →CD = 2 Vector
– 4

Zero Vector (Null Vector)

A vector whose magnitude is zero is called a zero vector. In a zero vector the initial and terminal points are

same. So, →AA= 0 , →BB = 0 are zero vectors.
0 0

Unit Vector

A vector having magnitude 1 (unit) is called a unit vector. A unit vector is generally denoted by ^a and read
as 'a cap'.

1

For example, 1 , 0 , 2 = etc. are the unit vectors.
0 – 1 3

2

If →AB = 3 is not a unit vector as →AB = 9 + 16 = 5
4

Types of Vectors ~ 223

Can you find a unit vector along →AB ?

Note: When a vector is divided by its own magnitude a unit vector along the vector is obtained.
→a
i.e. →a is a unit vector along →a .

Negative Vectors

Two vectors are said to be negative to each other if they have the same B
magnitude but opposite direction.
A B
If →AB displaces the point A to B. Then →BA displaces the point B to A.
A
So, →AB is negative vector of →BA and vice versa.

Hence, →AB = – →BA

If →AB = a then its negative vector is →BA and →BA = – →AB
b

=– a = – a
b – b

Parallel Vectors (Collinear Vectors)

Two vectors are said to be parallel if they are of same or opposite direction whatever may be their magnitudes.
That is two vectors →a and →b are said to be parallel if →a = k b→ where k is any positive or negative number.

D

PQ

B RS
AC

For example, if →AB = 1 →CD then →AB and →CD are parallel vectors in same direction. And if →PQ = – 2→RS
3

then →PQ and →RS are also parallel vectors in opposite direction.

Like Vectors

Two vectors are said to be like vectors if they have same directions →a →b →a →b
whatever may be their magnitudes.
i.e. Two vectors →a and →b are said to be like if →a = k →b for k > 0.

Unlike Vectors →u
→v
Two vectors are said to be unlike vectors if they have opposite directions
whatever may be their magnitudes.
i.e. Two vectors →u and →v are said to be unlike if →u = k →v for k < 0.

224 ~ Perfect Optional Mathematics Class 9

Perpendicular (Orthogonal) Vectors B
Q
Two vectors are said to be perpendicular or orthogonal vectors if AP

angle between them is 90º. The vectors →AB = 2 and →PQ = – 2 are A
4 1

orthogonal as direction of →AB = θ1 = tan– 1 4 = 63.43º = 63° (appr.)
2

direction of →PQ = θ2 = tan– 1 1 = 154º (appr.)
– 2

And angle between →AB and →PQ is 154º – 64º = 90º

Co-initial Vectors

T→OwAoavnedc→OtoBrs having the same initial point are called co-initial vectors.
are co-initial vectors.

Proper Vector O B

A vector whose magnitude is not zero is called a proper vector.

→AB = 3 , →PQ = 1 etc. are proper vectors.
5 0

Example 1:

If →AB displaces the point A(2, 5) to B(6, 1) and →PQ displaces the P(4, – 1) point to Q(8, – 5),
a. Find the components of →AB and →PQ.
b. Express →AB and →PQ as column vectors.
c. Find the magnitudes and directions of →AB and →PQ.

 a. x-component of →AB = 6 – 2 = 4 Vector

y-component of →AB = 1 – 5 = – 4 and

x-component of →PQ = 8 – 4 = 4

y-component of →PQ = – 5 + 1 = – 4

b. Hence, →AB = 4 and →PQ = 4 .
– 4 – 4
c. Magnitude of →AB =  →AB  = 16 + 16 = 4 2 units.

For direction of →AB is tan θ = – 4 = – 1 = – tan 45°.
4
As x-component +ve and y-component – ve of →AB ,

tan θ = tan(360° – 45°) = tan 315°

∴ θ = 315°
As →AB = →PQ , the magnitude and the direction of →PQ is same as that of →AB.

Types of Vectors ~ 225

Example 2:
If vector →PQ displaces the point P(2, – 1) to Q(5, 7), find a unit vector along →PQ .

 As vector →PQ displaces the point P(2, – 1) to Q(5, 7).

→PQ = 5–2
7+1

= 3
8

The magnitude of →PQ =  →PQ   = 32 + 82 = 9 + 64 = 73 units.

Hence, →PQ is not a unit vector.

The unit vector along →PQ is P^Q = →PQ
 →PQ  

= 1 3
73 8

= 1

73
8

73

Example 3:

If P, Q, R and S are four points with co-ordinates (2, – 2), (6, 4), (x, y) and (3, – 5) respectively such
that →PQ = →RS , find the co-ordinates of R.

 The co-ordinates of P and Q are (2, – 2) and (6, 4) respectively.

Hence, →PQ = 6–2 = 4
4+2 6

The co-ordinates of R and S are (x, y) and (3, – 5) respectively.

Hence, →RS = 3–x
– 5 – y

From question, we have →PQ = →RS

i.e. 4 = 3–x
6 – 5 – y

i.e. 4 = 3 – x and 6 = – 5 – y
y = – 11.
∴ x = – 1 and

226 ~ Perfect Optional Mathematics Class 9

EXERCISE : 6.3

1. a. Define unit vector with example.

b. Write the condition for →a and b→ to be parallel.

c. If →a = 2 and m is any scalar find m →a.
3

2. a. If →a = 2 and b→ = 4 , prove that →a || b→.
3 6
b. Write the formula to calculate the unit vector of →a.

c. If the length of a vector p→ = 6 is 10 units, find the unit vector of →p.
p

3. a. If →AB displaces the point A(– 7, – 3) to the point B(– 3, – 8), find the vector →AB ,  →AB   , the

direction of →AB and a unit vector along →AB .

b. If O→M = 2 and O→N = 6 are the position vectors of the points M and N respectively, find
– 3 2

 M→N  and direction of M→N.

4. a. If →a displaces the point A(4, 3) to the point B(5, 5) and →b displaces the point C(2, 0) to the point

D(1, – 2), prove that →a = – b→.

b. If A, B, C and D are four points with co-ordinates (2, 5), (2, 1), (5, 3) and (a, b) respectively. If
→AC = →BD , find values of a and b and the magnitude of →BD .

c. If M, N, A and B are four points with co-ordinates (– 1, 0), (3, 6), (x, y) and (2, – 4) respectively
such that M→N = →AB , find the co-ordinates of A and vectors M→B and →NA.

5. Prove that the following vectors are parallel. 1

a. →AB = – 5 and →CD = – 10 b. →PQ = 2 and →RS = 2 Vector
2 4 3 3

4

6. Prove that the following vectors are orthogonal (perpendicular).

a. M→N = 3 and →AB = 1 b. →AB = 1 and →PQ = – 3
– 1 3 3 2

7. →AB displaces the point A(2, 1) to B(– 1, 5). Express AB as a column vector. For each of the following

pair of points state whether →AB = – →PQ is true or not.

a. P(– 1, 2), Q(2, – 2) b. P(2, 3), Q(1, – 10)

8. If A(3, 1), B(2, 4) are two points then,

a. find →OA, →OB and →AB b. find  →OA  ,  →OB   and  →AB  

c. prove that vector →OA and →AB are perpendicular.



Types of Vectors ~ 227

6.4 Vector Operations

Vector operations means addition of vectors, subtraction of vectors, multiplication of vectors by a scalar
and product of vectors.

Addition of Vectors

Consider two vector →a and →b and place them making head of →a together with tail of →b, then the third vector
from the tail of →a to the head of →b will be obtained. The third vector is the sum of →a and →b and is denoted
by →a + →b . The sum is called the resultant of two vectors. Study the following three cases.

Case I, →a + →b
When →a and →b are like vectors then sum of →a and →b is vector in their →a →b
direction with the magnitude as the sum of their magnitudes.

Case II, →a – →b
When →a and →b are unlike vectors then the sum of →a and →b is vector in →a
the direction of the greater vector with the magnitude as the difference →b
of their magnitudes.

Case III, →a + →b →b
When →a and →b are non-parallel vectors then the sum of →a and →b is →a
a vector in the direction different from their directions with new
magnitude.

Triangle Law of Vector Addition A C
B
In the given figure, →AB displaces a point from A to B and →BC displaces
the same point from B to C then the total displacement of →AB and →BC
is the displacement from A to C. Which is denoted by →AC

Hence, →AB + →BC = →AC and is called Triangle Law of Vector Addition.

Definition

The triangle law of vector addition states that "If the magnitude and direction of two vectors are
represented by two sides of a triangle taken in order. The magnitude and direction of their sum is given
by the third side taken in reverse order."

228 ~ Perfect Optional Mathematics Class 9

Parallelogram Law of Vector Addition C D
A B
Let →AB = →a and →AC = →b be two co-initial vectors completing
parallelogram ABDC by taking AB and AC as adjacent sides and join
diagonal AD of parallelogram ABDC.
By the property of parallelogram, →CD = →AB = →a and →BD = →AC = →b

Now from ∆ABD,
A→D = →AB + →BD (From triangle law of vector addition)
i.e. A→D = →a + →b which is called Parallelogram Law of Vector Addition.

Definition

The parallelogram law of vector addition states that "If two adjacent sides of a parallelogram through
a point represents two vectors in magnitude and direction, their sum is given by the diagonal of the
parallelogram through the same point in magnitude and direction."

Polygon Law of Vector Addition

In polygon ABCDE, the sum of vectors represented by →AB , →BC , →CD D C
and →DE taken in order is equal to the vector represented by the vector E
→AE . Join AC and AD, Using vector addition,
AB
→AC = →AB + →BC

A→D = →AC + →CD = →AB + →BC + →CD Vector
→AE = A→D + →DE
∴ →AE = →AB + →BC + →CD + →DE

Definition

This law states that if a number of vectors can be represented in magnitude and direction by the sides of
polygon taken in order, their sum is given in magnitude and direction by the closing side of the polygon
taken in reverse order.

Addition of Two Vectors in Components Form

Suppose →a = x1 and b→ = x2 be two vectors then the sum of vectors →a + b→ is denoted by
y1 y2

→a + →b and is defined as →a + →b = x1 + x2 = x2 + x1
Alternatively, y1 y2 y2 + y1

Let →a = x1 and →b = x2 be two vectors. Draw →OA = →a = x1 and →OB = →b = x2 as shown in the figure.
y1 y2 y1 y2

Construct a parallelogram OACB with OA and OB as adjacent sides and join OC.

Vector Operations ~ 229

Draw BM⊥OX, AN⊥OX, CL⊥OX and AK⊥CL, DOBM @ ∆CAK and Y C(x2 + x1, y2 + y1)
OM = AK = NL = x2,
(x2, y2)
BM = CK = y2 and →AC = x2 = →OB .
y2 B

Hence, OL = x1 + x2 and CL = y1 + y2 OM A(x1, yK1)
Y' NL
∴ →OC = x2 + x1 X' X
y2 + y1

From the addition of directed line segments, →OC = →OA + →AB

Hence, →OC = →OA + →OB

or, →OC = x1 + x2
y1 y2

∴ →OC = x2 + x1
y2 + y1

Subtraction of Vectors

Let →AB = →a and →BC = →b be two vectors. The addition of →a and →b →a + →b C
is denoted by →AC . The difference between →a and →b is the sum of →a
→b

and – →b. Draw, BD such that BD = BC then A →a B

→BD = – →BC = – →b – →b

→AC = →AB + →BC = →a + →b D

A→D = →AB + →BD = →a + (– →b) = →a – →b (using triangle law of vector addition)

Note: If →a = x1 and →b = x2 then →a – →b = x2 – x1
y1 y2 y2 – y1

Multiplication of a Vector by a Scalar

Let →a = x1 , then 2→a = →a + →a = x1 + x1 →a
y1 y1 y1 →a
2→a
= x1 + y1 = 2x1 →b
y1 + y1 2y1
k→b
The vectors →a and 2→a are parallel and the magnitude of 2→a is double k→a
(k < 0)
the magnitude of →a . (k > 0)

If →b = x1 be a vector and k be a scalar k→b= kx1 .
y1 ky1

The vector k→b is a vector parallel to →b and its magnitude is k times the magnitude of →b.

230 ~ Perfect Optional Mathematics Class 9

Unit Vectors →i and →j

Let O be the origin and OX, OY be the positive X– axis and Y– axis Y

respectively. Then the unit vectors along X– axis and Y– axis are

denoted by →i and →j respectively are defined by →i = 1 and →i = 0
0 1

Every vector x can be represented by x→i + y→j and conversely. B
y
→j
Here, →i = 1 and →j = 0 X' X
0 1 O →i A

Y'

Then x→i + y→j = x 1 +y 0 Y
0 1 P(x, y)

= x + 0 = x y →j
0 y y

Conversely, x = x + 0 X' q → X
y 0 y xi
O M

=x 01 + y 0 = x→i + y→j Y'
1

Example 1:

If →a = 2 and →b = 2 , find →a + →b and →a – →b.
– 5 3

 Here →a = 2 and →b = 2 →a + →b →a
– 5 3 →a →b

→a + →b = 2 2 →a – →b
– 5 3
+

= 2+2 = 4 –→b Vector
– 5 + 3 – 2

→a – →b = 2 – 2
– 5 3

= 2–2 = 0
– 5 – 3 – 8

Example 2:

If →a = 2 and →b = – 4 , find 2→a – →b . Also find magnitude of 2→a – →b and unit vector along
– 3 2
2→a – →b.

 Here, →a = 2 and →b = – 4
– 3 2

2→a – →b = 2 2 – – 4 = 4 – – 4 = 4+4 = 8
– 3 2 – 6 2 – 6 – 2 – 8

Vector Operations ~ 231

Magnitude of 2→a – →b is |2→a – →b| = 82 + (– 8)2 = 8 2 units.

Unit vector along 2→a – →b is 1 →b | (2→a – →b ) = 812  8
|2→a – – 8

1

= 2
Example 3: –  1
2

If →a = – 2 and 2 →a + 3 →b = 2 , find →b .
– 4 3

 Here, →a = – 2 and let →b = x
– 4 y

We have, 2 →a + 3 →b = 2
3

or, 2 – 2 +3 x = 2
– 4 y 3

or, – 4 +3 x = 2
– 8 y 3

or, 3  x = 2 – – 4
y 3 – 8

or, 3  x = 2+4
y 3+8

or, yx = 1 6
3 11

or, xy = 2
11

3

∴ b→ = 2
11

3

Example 4:

In the figure, the diagonals PR and QS of a quadrilateral PQRS P S

intersect at O. If →OP = →a , O→Q = b→ , →OR = →c , →OS = →d prove →a →d
that →PQ + →QR + →RS + →SP = 0.
→b O →c

QR

 Here, →PO = – →OP = – →a , Q→O = – O→Q = – →b, →RO = – →OR = – →c and →SO = – →OS = – →d

232 ~ Perfect Optional Mathematics Class 9

→PQ = →PO + O→Q = – →a + →b = →b – →a
→QR = Q→O + →OR = – b→ + →c = →c – b→
→RS = →RO + →OS = – →c + →d = →d – →c
→SP = →SO + →OP = – →d + →a = →a – →d
and →PQ + →QR + →RS + →SP

= →b – →a + →c – →b + →d – →c + →a – →d
= 0 proved.

Example 5:

If →a = 2→i – 4→j and b→ = – 3 b→ + 2 →j , find the vector →c such that 2 →a – →b + 3 →c = 0.

 Here, we have →a = 2→i – 4→j , →b = – 3→i + 2→j and 2→a – →b + 3→c = 0

or, 2(2→i – 4→j ) – (– 3→i + 2→j ) + 3→c = 0
or, 4→i – 8→j + 3→i – 2→j + 3→c = 0
or, 3→c = – 7→i + 10→j

∴ →c = – 37 →i + 130 →j

Example 6:

PQRSTU is a regular hexagon, P U Vector
show →PR + →PS + →UP + →TP = 3 →PQ . T

Q

 PQRSTU is a regular hexagon and join UR. RS
PU
Then by the property of a regular hexagon, 2 →PQ = →UR .
Now, from ∆PRU, →UP + →PR = →UR QT
Again, from ∆PST, →TP + →PS = →TS
So, →PR + →PS + →UP + →TP RS

Vector Operations ~ 233

= →UP + →PR + →TP + →PS
= →UR + →TS
= 2 →PQ + →PQ  →TS = →PQ
= 3 →PQ

Hence, →PR + →PS + →UP + →TP = 3→PQ.

Example 7:

The positions vectors of the points P, Q, R and S are 2p→ + 4→r, 5p→ + 3 3 q→ + 4→r, – 2 3 q→ + →r and

2p→ + →r respectively. Show that PQ||RS and RS = 2 PQ.
3

 Let, O be the origin: R P Q
S
Then →OP = 2p→ + 4→r
O→Q = 5→p + 3 3 →q + 4→r
→OR = – 2 3 q→ + →r
→OS = 2→p + →r

i. →PQ = →PO + O→Q [By the triangle low of vector addition]
= O→Q – →OP
= 5→p + 3 3 →q + 4→r – 2→p + 4→r
=3p→ + 3 3 q→ = 3 p→ + 3 q→

ii. →RS = →OS – →OR [By the triangle low of vector subtraction]
= 2→p + →r – – 2 3 →q + →r
= 2p→ + 2 3 q→ = 2 →p + 3 →q

= 2 .3 →p + 3 →q = 2 →PQ
3 3

→RS = 2 →PQ
3

iii. RS||PQ [Being the result of (ii) in the form of →a = mb→]

Thus PQ||RS and RS = 23 PQ proved.

234 ~ Perfect Optional Mathematics Class 9

EXERCISE : 6.4A

1. a. What do you mean by closure property of triangle law of vector addition?

b. A vector →a = x and k be any scalar, find k→a.
y

2. a. If →a = 2 and b→ = 4 , find →a + b→.
3 5

b. If →a = 3 and b→ = 5 , find →a – b→.
4 6

c. If →a = (3, 5) and →b = (– 1, 3) write →a + →b in column vector.

3. a. If →a = 2 and →b = 1 , find the value of →a + 2b→.
4 – 2

b. If →a = –1 and 2→a – b→ = 1 , find →b.
4. a. If 4 of 3 and N are
7→i 3→j and – 5→i 2→j  M→N .
the position vectors M – + respectively, find

b. If →AB = 4 and →BC = 3 , find the values of →AB + →BC and  →AB + →BC  .
5 2

c. If →OA = – 5 and →OB = 6 , find →AB and  →AB  .
3 4

d. If →p = – 5 and →q = 3 , express →p – →q as a column vector and find its magnitude.
3 – 3

5. a. If →a = 2 and →b = 3  , then find  →a + 2→b  .
2 – 4

b. If →u = 2→i – 3→i and →v = →i + 2→j , find 2→u + →v and its magnitude.

6. a. If →a = 1 and →b = – 4 , obtain the following vectors.
3 2

i. →a – 2→b ii. 3→b – 2→a Vector

b. If →a = 2 , →b = 1 and →c = 1 , obtain the following in the form of column vectors.
– 3 – 2 3

i. →a + →b + →c ii. →a – →b + 2→c

7. a. If →p = 2→i – 4→j and 3→p – 2→q = 0, find the vector →q.

b. If →a = 2→i – 4→j and →b = 3→i + 2→j and →a + 2→b – →c = 0, find the vector →c .

8. a. If →a = 6 , →b = 2 , →c = 10 and →a + x→b = →c , find the value of the scalar x.
2 3 8

b. If →p = 2 , →q = –1 , →r = –1 , find k such that →p + k→q = →r .
1 2 7

Vector Operations ~ 235

EXERCISE : 6.4B

1. a. If →a = 2→i + 4→j and →b = →i – 3→j , find 2→a + 3→b in the form of column vector and find its magnitude

and direction.

b. If →a = 2 and →b = – 3 , express →a – →b in the form of column vector and also find its magnitude
1 4

and direction.

2. If →a = 1 , →b = – 3 and →c = 3 , find
2 4 6

a. →a – 2→b + 3→c as a column vector b. Magnitude of →a – 2→b + 3→c

c. Direction of →a – 2→b + 3→c

3. If the position vectors of two points P and Q are 3 and – 3 respectively, find the position vector of
7 4

M, the mid point of PQ by the method of

a. Addition of vectors b. Subtraction of vectors.
4. a.
Show that the point with the position vectors – 2 , 3 , 6 and 1 are the vertices of a
– 3 – 3 1 1

rhombus.

b. If the vectors →PQ = 2→i + 5→j and →QR = 3→i + 7→j and PQ and QR are the two adjacent sides of a

parallelogram PQRS, find the unit vector along diagonal PR.

5. a. If →AB = 3 and →CD = –4 , find the magnitude, direction and unit vector of →AB + →CD .
4 3

b. If →PQ = –2 and →RS = 3 then find the magnitude, direction and unit vector of →PQ + →RS .
7 –2

c. If →p = 4 , →q = –2 , find the magnitude, direction and unit vector of →p + 2→q .
3 3

d. If →a = –2 and →b = 3 , find the magnitude, direction and unit vector of 2→a – 2→b .
6 –1

6. a. If A→D = 2 , →AB = 4 , →BC = 2 and →AE = 9 , find P
4 1 5 – 2 MU
the column vectors for →AC, →DC, →EC , →BE from the given

diagram. CQ

b. If →QP = 3 , →UP = – 4 , P→M =– →RQ = 2 and D R
4 2 – 3 Q.N. 6(b) S

→RS = 8 , find column vectors for M→R, →RU and →US . B
– 2
A

Q.N. 6(a) E

236 ~ Perfect Optional Mathematics Class 9

7. a. In the given figure, PQRSTU is a regular hexagon, find U T

→UT, →QP and →US in terms of →a and →b. – 2→a

Two consecutive sides →AB and →BC of a regular hexagon P S
represent →a and →b respectively. Find →AC, A→D, →CD, →DE, Q
b. →a A F
E
→b R →a

→EF and →FA in terms of →a and →b. Q.N. 7(a) B

→b

CD

Q.N. 7(b)

8. a. If P and Q are the mid points of AB and AC respectively. A P
b. If A→Q = →a and →AP = →b, find →PQ, →BC, →PC and →BQ in PQ O
terms of →a and →b.
B Q.N. 8(a) C Q Q.N. 8(b) R
In the given figure, O is the middle point of PR. If →PQ = →p
and →QR = →q find →PR , →PO , →RO , Q→O

9. a. PQRS is a parallelogram whose diagonals intersect at O. PS
b. If →PO = →u and →OS = →v, find the →PQ and →QR in terms of
→u and →v. O P 2 →v S
Q Q.N. 9(a) R
PQRS is a trapezium in which PS//QR and PS:QR = 5:9
If →QP = →u and →PS = 5→v, find the vectors →QR , →PR , →QS →u
and →RS in terms of →u and →v.
Q Q.N. 9(b) R

10. a. In ∆ABC, if M be the mid point of BC, prove that : A Vector
→AB + →AC = 2A→M.
P

b. PQRST is a regular pentagon then prove that B M CQ T
→QR + →QP + →RS + →TS + →PT = 2 →QS .
Q.N. 10(a)
c. PQRSTU is a regular hexagon, show that
→PR + →PS + →UP + →TP = 3→PQ . RS
d. From the given regular polygon ABCDEF, prove that:
→AB + →BC + →CD + →DE + →EF = →AF . Q.N. 10(b)

PU

AF
QT

RS BE

Q.N. 10(c) CD

Q.N. 10(d)

Vector Operations ~ 237

e. In the given figure, →OL = →l , O→M = m→ and O→N = 3m→ – 2→l L
then prove that →LN = 3L→M M

ON

11. a. In the figure, →OA = →a, →OB = →b and →AB = 2→AP, find the B
b.
12. a. value of →OP in terms of →a and →b. →b P

In the figure, →OC = →u, O→D = →v and →CD = 3→CQ, find the O →a A →b D Q
value of O→Q in terms of →u and →v. Q.N. 11(a) O C
→a

Q.N. 11(b)

The position vectors of the points P, Q, R and S are →a + 3→b – 3→c, →a – →b, 5→c + 3→b and 8→c – →b

respectively, show that PQ//RS and RS = PQ.

b. The position vectors of the points A, B, C and D are 4→u + →v, 7→u – 2→v, 7→u + 2→v and 8→u + →v

13. a. respectivley, show that AB = 3CD and AB//CD. D SR
b. In the given figure, if A→D = 2→BC then prove that A
→AB + →CD = →BC. BC →c
c.
In the given figure, PQRS is a parallelogram. If →OP = →a, Q.N. 13(a) →a Q
O→Q = →b and →OR = →c, find →OS . O →b
P

Q.N. 13(b)

In the given figure, prove that : →AB – →CD – →AC = →DB. A D

d. In given figure, prove that A→O – →BO + →BC = →AC. AB
O
BC

Q.N. 13(c)

D Q.N. 13(d) C
A
e. In the given figure, ABCD is a parallelogram. If E is any poit B
on the diagonal AC then prove that →BE = →AE + →ED + →CE . E

 DC

238 ~ Perfect Optional Mathematics Class 9

7 TRANSFORMATION

Curriculum Contents Included Periods

- Introduction and types of tranformation by using coordinates (isometric 15
and non-isometric)

- Rotation through (i) ± 90° (ii) ± 180° (iii) ± 270° about (a, b) (graph and
formula)

- Reflection on X-axis, Y-axis, lines y = x, y = – x, x = a and y = b in graph
and using formula also

- Translation
- Enlargement and Reduction of geometrical figure about point (a, b) with

scale factor (formula or graph method)

Objectives

On completion of this unit, students will be able to:
find the reflected image of an object about the lines X-axis, Y-axis, y = x, y = – x, x = a and y = b
by drawing on graph and using formula also.

find the rotated image of an object through the angles ± 90°, ± 180°, ± 270° about the centre at
origin and (a, b) by drawing on graph and using formula also.

translate an object in given direction and magnitude.

enlarge or reduce an object on the basis of given scale factor and centre by drawing on graph and
using formula also.

Materials

T T' T' T' T Transformation
T' T T'
90° T
Reflection T Translation O O
Enlargement
Rotation Reduction

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Transformation 1 mark each 5 marks each Questons Marks
2 marks each 4 marks each
1 1 3 10
–1

 ~ 239

7 Transformation

Introduction

A transformation transforms or changes the position, shape or size of geometric figure. A figure before the
transformation is called an object and after the transformation is called an image. There are different types
of transformations.

Types of Transformations

Among various transformations, we study some of them in this level. In geometry a transformation T
maps every point A of the set of points of a plane figure to a unique point A' as the image of A under
transformation T. Each point has exactly one image point and each image point has exactly one pre-image
point under a transformation. So transformation is a one-to-one function.

Transformations

Identity Transformation

Isometric Transformation Translation

Rotation

Reflection

Non-Isometric Transformation

Enlargement

Reduction

Identity Transformation

If a transformation maps a point A to A itself, then it is called an identity transformation. It is denoted by I.
So, I(P) = P and the identity transformation leaves all points unchanged.

Isometric Transformation

The transformation in which the size of the figure remains same and only the position of the figure is
changed after transformation is called isometric transformation. Reflection, Rotation and Translation are
isometric transformations. Hence, the object and its image are congruent in isometric transformations.

240 ~ Perfect Optional Mathematics Class 9

Non-Isometric Transformation

The transformation in which the size of figure after transformation is changed is called non-isometric

transformation. Enlargement and Reduction are the non-isometric transformation. The object and the

image are similar in this transformation.

We will discuss about the following transformation in this chapter.

a. Reflection b. Rotation

c. Translation d. Enlargement (or Reduction)

Study the following figures for the general concepts of Reflection, Rotation, Translation, Enlargement and

Reduction.

O3

Reduction

Reflection O2 Object Enlargement

90° Translation
O3
Rotation
T T' T'
T' Transformation
90° T
Reflection T
Translation
Rotation

T' T
T T'

O O

Enlargement Reduction

Transformation ~ 241

7.1 Reflection

Introduction

Reflection is a transformation which has a mirror image about a line as a two way mirror. A line which plays
the role of two way mirror to give the image of given object is called the axis of reflection.

M
MM

P P' A A'

Q Object Image Q' D D'

Object Image

Object N Image R; R B C C' B'
N N

The following are the properties of reflection:
a. The object and the image under the reflection are congruent.
b. The object and its image will be at equal distances from the axis of reflection.
c. The line joining the object and its image will be perpendicular to the axis of reflection.

Example 1:

Find the image of a point P under the reflection over the line P A
AB.

 To have the image of the point P under the reflection over the line B A
P
AB, we perform the following steps.
O
Step 1: Draw PO⊥AB.
Step 2: Produce PO upto P' so that PO = OP'. B
The point P' is called the image of the point P under the reflection over P'
line AB.
Example 2: A
PS
Find the image of quadrilateral PQRS under the reflection over
the line AB.

Q

R' R
B

242 ~ Perfect Optional Mathematics Class 9

 Draw perpendicular from each vertex of quadrilateral PQRS on P A P'
Q S S' Q'
the axis of reflection AB and produce the perpendicular line upto
the point whose distance from the axis of reflection is equal to R' R A'
the distance of its object from the axis of reflection as in previous B
example and join P'Q'R'S' to get the required image.
A
Example 3:
The image of triangle ABC under the reflection is triangle C
A'B'C'. Find the line of reflection. C'

B

 Complete the following steps for the line of reflection. B' M
A
Step 1:
B C A'
Join any vertex of triangle ABC with its corresponding image. N C'

i.e. CC'. (We can join BB' or AA' instead) B'
Step 2:

Draw perpendicular bisector of CC', which is the line of axis of

reflection.
∴ MN is the line of axis of reflection.

Reflection Using Co-ordinates

The image of geometrical figures under the reflection over a line can easily be obtained with the help of

co-ordinates.

Reflection about X-axis Y

Let P(3, 2) be a point and X-axis be the axis of reflection. From P(3, 2) Transformation
point P draw PM⊥OX and produce PM upto P' so that PM = MP'
then the co-ordinates of point P' will be (3, – 2) as shown in figure.

X' OM X

P(3, – 2)

Y'

Hence, P'(3, – 2) is the image of the point P(3, 2) under the reflection about X-axis.

i.e. P(3, 2) X-axis P'(3, – 2)
Reflection

Reflection ~ 243

Let us see the following table of some points on the plane and their corresponding images under the
reflection about X-axis.

SN Points in the plane Corresponding images

1. A(3, 4) A'(3, – 4)

2. B(– 3, 2) B'(– 3, – 2)

3. C(– 4, 5) C'(– 4, – 5)

4. D(6, 3) D'(6, – 3)

5. M(a, b) M'(a – b)

From the above table we can see that the image of any point under the reflection about X-axis is obtained
by changing the sign of the y-coordinate of the given point. So, we can conclude for the image of a point
under the reflection about X-axis.

i.e. P(x, y) = X-axis P'(x, – y)
Reflection

Example 4:

Find the image of ∆ABC with vertices A(2, 3), B(5, 1) and C(3, 5) under the reflection about X-axis.

Draw the ∆ABC and ∆A'B'C' (image) on the same graph paper. YC
A
 As the vertices of given triangle ABC are A(2, 3), B(5, 1) and

C(3, 5), the vertices of the image ∆A'B'C' under the reflection

about X-axis can be obtained by using formula as follows:

We have, P(x, y) = X-axis P'(x, – y)
Reflection
B
Hence, A(2, 3) → A'(2, – 3) X' O B' X
A'
B(5, 1) → B'(5, – 1)

C(3, 5) → C' (3, – 5) Y' C'
Drawing ∆ABC and its image ∆A'B'C' on the same graph paper, we
have the figure as shown.

Note: We have the image ∆A'B'C' of ∆ABC either by using formula or by drawing it in graph paper whether
it is not given in question to draw in graph paper.

Reflection About Y-axis

Let A(4, – 2) be a point and Y-axis be the axis of reflection. From Y

point A draw AM⊥OY and produce it up to A' such that AM = MA', O
A'(– 4, – 2) M
then the co-ordinates of point A' will be (– 4, – 2) as shown in the

figure.

Hence A'(– 4, – 2) is the image of the point A(4, – 2) under the

reflection about Y-axis. X' X
A(4, – 2)
i.e. A(4, – 2) Y -axis A'(– 4, – 2)
Reflection

Y'

244 ~ Perfect Optional Mathematics Class 9

Let us see the following table of some points on the plane and their corresponding images under the
reflection about Y-axis.

SN Points in the plane Corresponding images

1. P(– 2, 3) P'(2, 3)

2. Q(4, 6) Q'(– 4, 6)

3. R(3, – 5) R'(– 3, – 5)

4. S(– 2, – 4) S'(2, – 4)

5. N(a, b) N'(– a, b)

From the above table we can see that the image of any point under the reflection about Y-axis is changed

by the sign of its x-coordinate from the given point. So, we can conclude for the image of a point under the

reflection about Y-axis.

i.e. P(x, y) = Y-axis P'(– x, y)
Reflection

Example 5:

Find the image of quadrilateral PQRS with vertices P(– 4, 2), Q(– 3, 4), R(1, 5) and S(– 2, 1) under

the reflection about Y-axis. Draw quadrilateral PQRS and its image P'Q'R'S' on the same graph

paper.

 As the vertices of given quadrilateral are P(– 4, 2), Q(– 3, 4), Y

R(1, 5) and S(– 2, 1), the vertices of the image P'Q'R'S' can be

obtained by using formula (or by drawing on graph paper) as R' R
Q Q'
follows:

We have, P(x, y) Y-axis P'(– x, y)
Reflection

Hence, P(– 4, 2) → P'(4, 2) P S S' P'
Q(– 3, 4) → Q'(3, 4) X' O X
R(1, 5) → R'(– 1, 5)

S(– 2, 1) → S'(2, 1)

Drawing quadrilateral PQRS and its image P'Q'R'S' on the same Y'
graph paper, we have the figure as shown.

Reflection About the Line y = x Y Transformation

Let ∆ABC be a triangle with vertices A(6, 4), B(5, 1) and C(3, 2) B' y=x X
and the line y = x be the axis of reflection. The line y = x is a line
bisecting the angle between X-axis and Y-axis lying in the first C' A'
and the third quadrant. Draw a perpendicular AM on the axis of C M
reflection and produce it upto point A' so that AM = MA' then the A
co-ordinates of the point A' will be (4, 6) as shown in the figure. X' O
In the similar way the co-ordinates of B' and C' can be found as B
(1, 5) and (2, 3) respectively as shown in the figure.
Y'

Hence, ∆A'B'C' is the image of ∆ABC under the reflection about the line y = x.

Reflection ~ 245

Let us see the following table of some points on the plane and their corresponding images under the
reflection about the line y = x.

SN Points in the plane Corresponding images

1. E(2, 4) E'(4, 2)

2. F(– 3, 1) F'(1, – 3)

3. G(– 2, – 3) G'(– 3, – 2)

4. H(5, – 2) H'(– 2, 5)

5. Q(a, b) Q'(b, a)

From the above table we can see that the image of any point under the reflection about the line y = x is
obtained by interchanging the x-coordinate and y-coordinate of the given point. So, we can conclude for the
image of a point under the reflection about the line y = x.

i.e. P(x, y) = y=x P'(y, x)
Reflection

Reflection About the Line y = – x Y

Let P(– 2, 5) be a point and the line y = – x be the axis of reflection. P(– 2, 5)
The line y = – x is a line bisecting the angle between the X-axis and
the Y-axis lying in the second and the fourth quadrant. From point P M
draw PM perpendicular on the line

y = – x and produce it upto P' so that PM = MP', the co-ordinates of P'(– 5, 2)

P' is (– 5, 2) as shown in the figure. X' O X

Hence, P'(– 5, 2) is the image of the point P(– 2, 5) under the Y'
reflection about the line y = – x.

i.e. P(– 2, 5) y = – x P'(– 5, 2)
Reflection

Let us see the following table of some points on the plane and their corresponding images under the
reflection about the line y = – x.

SN Points in the plane Corresponding images

1. A(– 3, 5) A'(– 5, 3)

2. B(2, 6) B'(– 6, – 2)

3. C(– 4, – 3) C'(3, 4)
4. D(3, – 2) D'(2, – 3)
5. E(a, b) E'(– b, – a)

From the above table, we can see that the image of any point under the reflection about the line y = – x is
obtained by interchanging the x co-ordinate and y co-ordinate with opposite signs of the given point. So,

246 ~ Perfect Optional Mathematics Class 9